That seems like a negative for your theory, right there. All of special relativity can be derived in a chapter or two of a standard undergraduate textbook. That seems much simpler than however you did it.That's a loaded question for me. I believe my list of equations is the result of only manipulating my two formulas but I'd have to go through years of derivations to confirm that's true.
Okay. That's interesting. No time dilation in your version.My math also shows time does not slow in time dilation, it has to do with relativity of simultaneity depending on when the stop watches are started from each perspective. This is just a math exercise, philosophy can be debated but math can't be, it's either right or wrong.
Never mind, for now.Yes, I'll go look it up and write it out if you insist but I won't be using tex to save me time.
What's the point of having the formula, then, if it's not useful for anything?I don't know if there's any application.
What do you mean when you say "c has 2 velocity components"? c is a constant, isn't it? The speed of light in vacuum?It was an example that if c has 2 velocity components in the main equation then those two velocities are subject to the velocity combo equation which is the other main equation.
I answered these questions twice already. My math is much simpler, you haven't seen the full math yet. It does not disagree with the experimental results of relativity as I've said twice alreadyThat seems like a negative for your theory, right there. All of special relativity can be derived in a chapter or two of a standard undergraduate textbook. That seems much simpler than however you did it.
I should also ask: does your theory disagree with special relativity, in terms of any testable predictions? Or is it just a reformulation of special relativity?
Never said that. I said there's no time slowing. I emphasize use of relativity of simultaneity.No time dilation in your version.
I've already shown your example in my spacetime diagram.Is your answer the same as the answer from special relativity?
I already showed the derivation in my next post where I recopied your original 2 questions.Never mind, for now.
I already explained the usefulness.What's the point of having the formula, then, if it's not useful for anything?
Fully explained. You're just not reading or understanding what I writeWhat do you mean when you say "c has 2 velocity components"?
example?
Sure, but that's not your "enhanced" velocity addition formula. To be clear, I'm talking about this one:c2=v2+v2t c^2 = v^2 +v_t^2 so c^2=3/5^2 +4/5^2 no?
Yes.Can you read spacetime diagrams?
So $v_t$ is a dimensionless number.$$v_t$$=c/Y=ct′/ct is the velocity or rate of time through time.
I have no idea what that means. Are you just saying that no observer sees clocks in his own frame as ticking at anything other than the "normal rate"? That would just be the same as saying that all inertial frames are equivalent, approximately. Is that what you're saying, or are you saying something different?Now all clocks in their frames beat at the normal rate c through time.
Not in relativity. In your version of relativity. Do you have a name for your theory, by the way?There are 3 types of rates through time in relativity.
I don't know what you mean by an "imbalance of relative velocities". How do we "balance" relative velocities? By "balance" are you just saying that two relative velocities are different from one another, or are you saying something else?One the permanent slowing of time during an imbalance of relative velocities when one participant initiates a change which takes time for the other to register the change....
..., another is the Doppler Shift Ratio which is partly an apparent rate caused mostly by the rate of info delay from a moving clock.
Is $v_t$ relative, then?The third is $$v_t$$ which satisfies the main equation so that the faster you are observed through space, the slower your time rate is observed.
Please correct me if I'm wrong here.For example if v=3/5c, $$v_t$$=4/5, DSR = .5c and if the participant was to turn around after 3 ly, she would age 2 yrs less when the two re-unite at Earth (i.e. the twin paradox.)
Or maybe it's the other way around. It doesn't really matter, either way, I think.Before I continue, consider this very important fact: space (and substances like water or glass) house an electromagnet medium with permittivity and permeability properties that define the speed of the electromagnetic wave that propagates using that medium.
I don't think it's very helpful to think of a vacuum as being like a material medium (like water, for example). A vacuum is empty space.The important difference is material has no relative velocity to a vacuum's electromagnetic medium. Why? Because c travels through its medium and if you were able to push the medium you'd be able to add that push to c but you can't because c is the speed limit. If you blasted a vacuum bottle into space, the relative velocity of the bottle to the vacuum it contains is the same as the bottle's relative velocity to the vacuum it's speeding through.
That's a bit muddled. Light certainly has a velocity relative to the bottle. If you rode along on the bottle, you'd measure the speed of light to be c, relative to the bottle. The problem is that you can't ride along on a beam of light, so it doesn't make much sense to say the bottle has a velocity of c relative to the light. I guess if you wanted to strain things, you might say that everything has velocity of c relative to light. That would be better than saying everything has "no velocity" relative to light. But perhaps we're just saying the same thing in different ways?Since light is an EM wave travelling through an EM medium, the bottle also has no velocity relative to light.
You'd need to be careful to use the correct relativistic velocity additional formula, or you'd get the wrong answer.The Fizeau experiment suggests that the EM medium of water limits the speed of light to .75c. This means if you blasted a water bottle into space, the velocity of the bottle could be added to the medium and hence to the light that travels through that medium. It would act as any wave through a medium moving relative to an outside observer. The medium has a relative velocity to an outside observer and that would be added to the speed of light in that medium so long as that addition does not violate c as the speed limit.
I don't really understand this. For example, on a Minkowski diagram, a constant relativity velocity in the frame of the diagram can be depicted as a straight line on the graph.Anyway, relativity uses spacetime diagrams (Minkowski, Epstein or Loedel) to try to depict relative velocity. However they are incomplete without considering each perspective as stationary so two diagrams are needed to depict relative velocity.
Adding a third observer to a scenario with two observers does not make the situation unphysical.Although a spacetime diagram is supposedly about the two ships, Earth is also a participant and so is this non-physical cartesian coordinate background which is the true reference frame in empty space.
Okay. For some reason you want to use a reference frame in which the Earth is moving. I don't know what you gain from doing that, exactly. Am I missing something?The Earth is travelling at what I call the Loedel or half speed $$v_h$$ velocity through space.
Wouldn't that be zero velocity? If there are two velocities $\pm v$, then the "average" of the two is zero, not $v/2$.Wiki defines it as an intermediate perspective between two equal velocities of opposite sign.
How long have you been at this? You previously said 10 years or something, didn't you? It beggars belief that in 10 years you haven't found anybody who is familiar with Minkowski diagrams.(I'm paraphrasing because this is difficult to find on Wiki.) It is completely ignored by relativity but is extremely important to this math exercise. If you want I can provide how all this post looks using spacetime diagrams but I've found no one knows how to read them so they get intimidated or suspicious I'm trying to pull a fast one on them. Plus they take a lot of work.
I don't understand what you're saying. When should the distance between Bob and Alice be "the same"? The same as what other distance (i.e. what two distances are you comparing)? According to which observer? At what time? Using whose clock and whose rulers?So a prime rule of relativity is the principle of relativity. This carries over into how relative velocity is depicted in spacetime diagrams, all the physical parameters should be the same no matter how it's drawn. But I can see at least 2 disturbing differences between how the Loedel diagram differs from the Minkowski depiction:
1. The distance separation between Bob and Alice should be the same. At t'=2, it's 1.5 ly in the Md but only 1.414 ly in the Ld.
Yes, you're right. It's a consequence of the Lorentz transformation between the two different observer frames. Both time dilation and length contraction effects are seen.It's because the stationary frames are different, one is the Earth and the other is a blank empty space. The time units of the blank space are different from Earth's time units so the distance units will also differ. The Cartesian grid is square and it's a different unit size in the two depictions.
Not a problem. If the two signals are both travelling at c, then one will travel twice as far in twice the time.2. The light signals are not the same length even though they represent the same velocity of c. The pink light signal travels twice as far in twice the time (as the yellow) chasing Alice speeding away from it.
You're already having to compensate, by having to mark off different t and t' coordinates on the various worldlines you've drawn.If I'm going to represent both perspectives on a single spacetime diagram, I'm going to have to compensate for this apparent imbalance in symmetry when depicted in a single spacetime diagram.
As long as the light travels at c in all frames of reference, there's no problem in terms of relativity.This actually has huge implications. Imagine a football thrown at 10m/s relative to the field. The speed and direction of the catcher determine the relative velocity of the catcher to the football but the football relative to the field remains at 10m/s. Normally the relative velocity of catcher to football would decrease if he's running away and increase if he's going toward it. If the football was light, his movement towards the light can't have a relative velocity to c greater than c. There's no danger of that happening if he's moving away yet his relative velocity to c is still c as shown in the velocity combo formula. The reason it doesn't is because the field is a vacuum and his movement has no relative velocity to the vacuum in any direction. However, in the spacetime diagram, Bob's pink light to Alice chases Alice. It takes that light more time over greater distance to catch up which is the definition of decreased relative velocity except Alice can have no relative velocity to light's medium otherwise her relative velocity to c in the opposite direction would exceed c.
But water isn't the same as vacuum. There's no requirement that a football travel at the same speed in all frames in water (or even in a vacuum for that matter). There's also no requirement that light should travel at the same speed in water in all frames.If the field was water, the ball could only travel 7.5m/s through that medium which would allow the motion of the catcher relative to the medium to affect his relative velocity to the ball.
So you're basically asking me to show you how to plug in numbers from my tables into a formula and you meant to say v=3/5c and u=4/5c and you still have no idea what u_t and v_t are and can't find them in the headers of my tables? Ok if v=3/5, v_t = 4/5 and if u=4/5, u_t=3/5. I didn't even need to look at the tables. Should I take a video of me punching numbers into a calculator? I don't think you're being serious so really if you don't want to put any effort into trying to understand anything I've written, then we should part ways and consider this thread one of the great mysteries in life you have no time for. Like I said, I'm busy.Sure, but that's not your "enhanced" velocity addition formula. To be clear, I'm talking about this one:
c2=((v+u)2+v2tu2t)/(1+vu/c2)2c2=((v+u)2+vt2ut2)/(1+vu/c2)2c^2= ((v+u)^2+ v_t^2u_t^2) / ( 1 + vu/c^2)^2
What are vvv, uuu, vtvtv_t and ututu_t?
I asked you to give a few examples of the application of that formula.
IfDSR_v= Y_u/Y_w where w is the combo of v and u
as u approaches c, both Y_u and Y_w approach infinity but the infinites cancel out in their ratio which approaches a finite value of DSR_v.
I now understand that $v_t$ is just related to whatever $v$ is by $c^2=v^2+v_t^2$. I don't understand why anybody would want to use $v_t$ for anything, but that's a separate issue. I can certainly calculate it from the meaningful velocity $v$ if I need it.So you're basically asking me to show you how to plug in numbers from my tables into a formula and you meant to say v=3/5c and u=4/5c and you still have no idea what u_t and v_t are and can't find them in the headers of my tables? Ok if v=3/5, v_t = 4/5 and if u=4/5, u_t=3/5.
No need. Please assume I am able to use a calculator.Should I take a video of me punching numbers into a calculator?
I have asked you some specific questions in my posts above and I have posted many specific comments about what you have written. I have also pointed out several apparent errors in your work. Why are you ignoring the posts where I did those things?I don't think you're being serious so really if you don't want to put any effort into trying to understand anything I've written, then we should part ways and consider this thread one of the great mysteries in life you have no time for. Like I said, I'm busy.
I listed a bunch of forms of this equation including Y=c/sqrt(c2-v2) because v_t =c/Y. So you're saying all the other forms are also dimensionally incorrect?That equation is dimensionally incorrect if vtvtv_t has different units from vvv and ccc.
I've found out why Ralf has come back here: https://www.scienceforums.com/topic/37634-ralfativity/?tab=comments#comment-390894I now understand that $v_t$ is just related to whatever $v$ is by $c^2=v^2+v_t^2$. I don't understand why anybody would want to use $v_t$ for anything, but that's a separate issue. I can certainly calculate it from the meaningful velocity $v$ if I need it.
I don't understand why you keep randomly adding and dropping factors of 'c', though.
Do $v$ and $v_t$ have the same units (metres per second, say), or not?
If not, then your equation $c^2 = v^2 + v_t^2$ must be incorrect. Do you agree?
If, on the other hand, $v_t$ does have the same units as $v$ and $c$, after all, then it cannot be the case that
$v_t=\frac{ct'}{ct}$,
as you previously stated, because that would be a dimensionless (unitless) quantity.
No need. Please assume I am able to use a calculator.
I have asked you some specific questions in my posts above and I have posted many specific comments about what you have written. I have also pointed out several apparent errors in your work. Why are you ignoring the posts where I did those things?
I have also asked you to clarify some of your definitions so I can better understand you. Will you do that, or not?
If you don't want to put any effort into explaining your own work, then we should part ways and consider your thread to be one of the great wastes of time in life. I'm busy, too, so if you're not serious we can call this quits now.
I think you need to decide whether you came here for a discussion, or just to publish your ideas. This is a discussion forum, not your blog.
The Y you give here is a dimensionless constant, according to its formula.I listed a bunch of forms of this equation including Y=c/sqrt(c2-v2) because v_t =c/Y. So you're saying all the other forms are also dimensionally incorrect?
Well that's a pity. Here you are, never having found somebody who can understand a Minkowski diagram in 10 years, and you decide to ignore the first person you meet who does understand it.I'm not even going to read or answer anything else from you.
Don't be silly. If you answered my questions, obviously there would be no point in my repeating them.You keep asking the same questions that I've already answered.
That's a pity, because I could probably help you and save you some time.No more time for you.
Okay.However Janus58 is a guy who knows his stuff so I'd be happy to talk to him instead.
The Y you give here is a dimensionless constant, according to its formula.
In the other formula you give here, for v_t, it seems that v_t must have the same dimensions as c, because Y is dimensionless. If c is a velocity in metres per second, say, then according to that equation v_t must also be a velocity in metres per second.
Your "main equation", then, is fine as long as c, v and v_t all have the same dimensions. In that case, I assume you've just made some careless errors every time you have used an example where v_t has no units.
The only other problem we need to sort out is what your error is in the equation:
$v_t = ct' / ct = t'/t$, which obviously has v_t as a dimensionless quantity rather than a velocity.
Is that equation supposed to be correct, or do you want to amend it in light of the above?